Find the area of a triangle ABC, where AB = 6.8m, BC = 9.5m and AC = 10.2m.
The cosine of a triangle is given by;
`cosA = (b^2+c^2-a^2)/(2bc)`
A is the angle between AB and AC.
a = Opposite leg to the angle A.(leg BC) = 9.5cm
b = Opposite leg to the angle B.(leg AC) = 10.2cm
c = Opposite leg to the angle C.(leg AB) = 6.8cm
`cosA = (10.2^2+6.8^2-9.5^2)/(2*10.2*6.8) = 0.433`
We know that ` sin^2A+cos^2A = 1`
`sinA = sqrt(1-cos^2A) = sqrt(1-0.433^2) = 0.813`
If we assume that the perpendicular drawn from point B will meet AC at point D;
`AD = ABsinA = 6.8*0.813 = 5.528`
Area of triangle
= 1/2(base length*perpendicular height)
So the area of the triangle is `28.193cm^2`
To find the area of the triangle, given the length of the three sides, we may use the Heron's Formula.
`A = sqrt(s(s-a)(s-b)(s-c))`
where A - Area of triangle
a, b, c - length of the sides of the triangle
s - semi-perimeter, `s = (a+b+c)/2`
So, let a =AB=6.8m, b = BC=9.5m and c=AC=10.2m.
Then, solve for s.
`s = (a+b+c)/2 = (6.8 + 9.5+10.2)/2 = 26.5/2= 13.25`
And, substitute values of , b, c and s to the formula of area of triangle.
Hence, the area of triangle is `31.26m^2` .